Showing time changed brownian motion is martingale. Let $W$ be a one dimensional Brownian motion and define, 
$$
X_t=W_{(\text{exp}(\beta t)-1)}\\
\hat{W}_t=\frac{1}{\sqrt{\beta}}\int_0^te^{-\frac{\beta s}{2}}dX_s
$$
Show that $\hat{W}_t$ is a local martingale in its natural filtration and compute its quadratic variation.
To show that it's a local martingale, is it just straight forward differentiation on $\hat{W}_t$ like below, 
$$
d\hat{W}_t=\frac{1}{\sqrt{\beta}}e^{-\frac{\beta t}{2}}dW_{(\text{exp}(\beta t)-1)}
$$
 A: As @muaddib pointed out, you have simply rewritten the definition of $\hat{W}_t$ - but this doesn't show that $(\hat{W}_t)_{t \geq 0}$ is a martingale.
Hints:


*

*Show that $(X_t)_{t \geq 0}$ is a martingale with respect to its canonical filtration $$\mathcal{F}_t := \sigma(X_s; s \leq t) = \sigma(W_s; s \leq e^{\beta t}-1).$$

*Conclude that $(\hat{W}_t)_{t \geq 0}$ is a martingale with respect to $(\mathcal{F}_t)_{t \geq 0}$.

*Using the tower property show that $(\hat{W}_t)_{t \geq 0}$ is also a martingale with respect to its natural filtration.

A: I came up with a solution involving properties of time change as follows...
Let $\theta(t)=\int_0^t\xi^2ds$, which we assume is finite. 
Then the stochastic integral $Y_t=\int_0^t\xi dW_s$ exists, and $Y_t$ is a Gaussian process with independent increments. 
The variance of $Y_t-Y_s$ is given by,
$$
\begin{align}
Var\left[Y_t-Y_s\right]&=\mathbb{E}\left[\left(Y_t-Y_s\right)^2\right]-\mathbb{E}\left[Y_t-Y_s\right]^2\\
&=\mathbb{E}\left[\left(\int_0^t \xi dW_s - \int_0^s\xi dW_u \right)^2\right]-0\\
&=\mathbb{E}\left[\left(\int_s^t \xi dW_u  \right)^2\right]\\
&=\mathbb{E}\left[\left(\int_s^t \xi^2du  \right)^2\right]\\
&=\theta(t)-\theta(s)\\
&=\mathbb{E}\left[\left( W_{\theta(t)}-W_{\theta(s)}  \right)^2\right]\\
\end{align}
$$
So $Y_t$ has the same distribution as time changed Brownian Motion $W_{\theta(t)}$. From the definition of $Y_t$ we have,
$$
Y_t=\int_0^t\xi dW_s=\int_0^t\sqrt{\theta'(s)}dB_s \sim W_{\theta(t)}
$$
For our problem, $ \theta(t) = e^{\beta t}-1$ , and $\sqrt{\theta'(t)}=\sqrt{\beta}e^{\frac{\beta t}{2}}$
From this we finally have,
$$
\hat{W}_t=\frac{1}{\sqrt{\beta}}\int_0^t e^{-\frac{bs}{2}}\sqrt{\beta} e^{\frac{bs}{2}}dW_s=\int_0^t dW_s
$$
Which is a martingale. 
For quadratic variation, we know that $\langle \hat{W}\rangle_t$ is the process such that $ \hat{W}_t^2 - \langle W\rangle_t$ is a martingale. Therefore, 
$$
\begin{align}
\mathbb{E}\left[ \hat{W}_t^2 - \langle \hat{W}\rangle_t \right] &= 0\\
\Rightarrow\mathbb{E}\left[ \langle \hat{W}\rangle_t \right] &= \mathbb{E}\left[ \hat{W}_t^2 \right]\\
 &= \mathbb{E}\left[ \left(\int_0^t dW_s \right)^2\right]\\
 &= \mathbb{E}\left[ \left(\int_0^t 1^2 ds \right)^2\right]\\
\Rightarrow \langle \hat{W}\rangle_t = t
\end{align}
$$
This shows that $\langle \hat{W}\rangle_t $ is a Brownian Motion in its natural filtration.
A: @saz i tried out the following according to your hints, 
Let $\mathcal{F}_t:= \sigma(X_s;s\le t)$, then we have, 
$$
\begin{align}
\mathbb{E}[X_t|\mathcal{F}_s]&=\mathbb{E}[X_t-X_s+X_s|\mathcal{F}_s]\\
&=\mathbb{E}[X_t-X_s\mathcal|{F}_s] + \mathbb{E}[X_s|\mathcal{F}_s]\\
&=X_s
\end{align}
$$
So $X_t$ is a martingale under its natural filtration, and $\hat{W}_t $ is a martingale with respect to $\mathcal{F}_t$.
But I'm not sure how to what is the tower property part you have in mind. Can you kindly elaborate?
